Let $X$ be a regular integral noetherian scheme of dimension 2 and let $D$ be a simple normal crossings divisor in $X$.

EDIT: Let $U = X-D$.

Consider a finite etale morphism $V\longrightarrow U$ with $V$ connected. Let $\pi:Y\longrightarrow X$ be the normalization of $X$ in the function field of $V$. So $Y$ is a $2$-dimensional normal noetherian scheme and $\pi$ is finite.

I suppose $U = X - D$. Are you using EGA IV$_4$, sec. 19 on "c.i. morphisms"? In view of flatness, then amounts fiber rings being ci rings. As noted after Theorem 23.6 of Matsumura's "Commutative Ring Theory", by using Andre homology it follows from the ci property of $X$ that it is equivalent that $Y$ be a ci ring. So the question concerns the finitely many non-regular pts of $Y$ lying over $D$. In view of Abhyankar's Lemma, for a counterexample something non-tame will need to happen over one of the generic points of $D$ (as otherwise $Y$ would even be regular). Perhaps you knew all this...
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BCnrdOct 3 '10 at 16:20

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Brian, a tame cover can be singular at a point lying over a singularity of $D$ (think of $z^2=xy$). I have a feeling that if you take a tame cyclic cover of the form $z^n=xy^a$ then its normalisation might well not be l.c.i. - perhaps even $(n,a)=(7,3)$ will do the trick, though it is too late at night for me to be certain.
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Tony SchollOct 4 '10 at 1:00

Tony, I think you're right. I think the singularities of Y will occur in the inverse image of the singular locus of D. In particular, the morphism will be etale above D-Sing(D). I can prove these things quite easily when we work over the complex numbers using Grauert-Remmert. But I wouldn't know how to do it for surfaces over any algebraically closed field of char 0 or even more interesting: any arithmetic surface. The property of being an lci will thus intuitively not be fulfilled at the points above D^{sing}...
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Ariyan JavanpeykarOct 4 '10 at 6:36

When $X=Spec A$ is regular and local, and $D$ is a n.c.d. defined by $f_1...f_r=0$, then, up to étale cover, $Y$ is the quotient of $Z:=Spec A[T_1,...,T_r]/(T_1^{n_1}-f_1,..., T_r^{n_r}-f_r)$ by a subgroup of $\mu_{n_1}\times ... \times \mu_{n_r}$, where $n_i$ is the ramification index above the component $f_i=0$. This is a consequence of Ahbyankar's lemma, see SGA1, XIII, 5.3.0 (and extend $U$ to $X$). When $r=1$ (for $D-Sing(D)$), then $Y=Z$ is regular hence l.c.i. over $X$. If $r\ge 2$, $Z$ is still regular, $Y$ is dominated by $Z$, but not necessarily l.c.i. over $X$ as showed Tony.
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Qing LiuOct 4 '10 at 22:02

2 Answers
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Let $k$ be a field of characteristic prime to $n$ and $\zeta\in k$ a primitive $n$-th root of unity. Let $a\in \mathbb{Z}$ and let $Y$ be the quotient of $\mathop{\rm{Spec}}k[u,v]$ by the automorphism $(u,v)\mapsto (\zeta u,\zeta^a v)$.

Then $Y$ is a tame cyclic cover of $X=\mathop{\rm{Spec}}k[u^n,v^n]$ etale away from $D=\{uv=0\}$. Typically $Y$ is not an lci. For example, if $(n,a)=(3,1)$ then
$$Y=\mathop{\rm{Spec}} k[u^3,v^3,uv^2,u^2v]=\mathop{\rm{Spec}}k[x,y,z,t]/(xy-zt,xz-t^2,yt-z^2).$$
If you allow wild ramification I suspect you can produce an example in which $D$ is even a smooth divisor.

Yes, Y is not a lci in this example because the closed immersion of Y in affine 4-space is not regular. Just to get it right, though...This also implies that the morphism pi from Y to X is not an lci?
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Ariyan JavanpeykarOct 4 '10 at 13:11

Good. This is exactly the local model of my last example (the cone over a rational normal cubic in $\mathbb{P}^3$).
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Francesco PolizziOct 4 '10 at 13:28

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@Francesco - thank you! Regarding wild ramification, the cover given by $t^p-y^3t+xy^2=0$ is etale away from $y=0$. For $p=5$ and $7$ I computed its normalisation in SINGULAR - the fibre at $x=y=0$ is not a complete intersection.
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Tony SchollOct 4 '10 at 14:57

I think the answer is in general no, because of the following example.

Take $10$ lines in $\mathbb{P}^2$ all passing through a point $p$, and let $D$ be their union. There exist a covering $Y \to \mathbb{P}^2$ of degree $5$ branched on $D$, and $p$ is the only point of total ramification. The only singularity of $Y$ is the unique preimage $v$ of $p$.

In order to understand the singularity of $Y$ at $v$, let us consider the blow-up $\widehat{\mathbb{P}}$ of $\mathbb{P}^2$ at the point $p$; we obtain an exceptional curve $E$, and by the Riemann-Hurwitz formula we see that the preimage of $E$ in the induced cover $\widehat{Y} \to \widehat{\mathbb{P}}$ is an elliptic curve with self-intersection $(-5)$.

From this we recognize that $Y$ is nothing but the elliptic normal cone in $\mathbb{P}^5$ (that is, the cone over the elliptic normal curve of degree $5$ in $\mathbb{P}^4$) and $f \colon Y \to \mathbb{P}^2$ is a general projection to $\mathbb{P}^2$.

Now, Saito (Einfach-elliptische Singularitaten, 1974) proved that the singularity at the vertex of an elliptic normal cone (a so-called simple elliptic singularity) is a complete intersection if and only if the degree of the cone is $\leq 4$.

In this case the degree is $5$, so $Y$ is not a local complete intersection.

All the stuff about the covering obviously works in any characteristic (well, maybe different from $5$), but I think Saito's result requires the ground field to be $\mathbb{C}$, so I do not know whether this works also in the case of positive characteristic.

EDIT As pointed out by Angelo, in this example $D$ is not simple normal crossing, so it
does not really answer the question. Anyway, let me try to give another example which works with normal crossing $D$.

Take the product $X=E \times F$ of two elliptic curves, and consider the divisor $D=E_1+E_2+F_1+F_2$, where the notation is obvious. Now $D$ has only nodal singularities, so it is simple normal crossing. Moreover, by using Pardini's construction explained in "Abelian covers of algebraic varieties", it is possible to construct a $\mathbb{Z}_3$-cover $Y \to X$ whose singularities are

two rational double points of type $A_2$ and

two points of type $1/3(1,1)$

(these singularities obviously lie over the four singular points of $D$).

The points of type $1/3(1,1)$ are locally analitycally a cone over a twisted cubic curve in $\mathbb{P}^3$, so they are not complete intersection singularities.

I am afraid that your $D$ is not a normal crossing divisor. In characteristic 0 the scheme $Y$ should always be l.c.i., because of Abhyankar's Lemma.
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AngeloOct 4 '10 at 9:19

@Angelo: I'm a bit confused. Does your statement imply that if X is fibered over a Dedekind scheme B with function field K(B) of characteristic zero, the morphism \pi is a local complete intersection (in the sense of Liu Section 4.6.2)?
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Ariyan JavanpeykarOct 4 '10 at 9:27

@Angelo: I checked the construction and I found no errors. Maybe I'm missing something, but I do not understand exactly what. How do you exactly use Abhyankar's lemma to prove that $Y$ must be lci? As far as I know, it says that I can get rid of (tame) ramification by taking a finite, branched cover $Z \to \mathbb{P}^2$...
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Francesco PolizziOct 4 '10 at 9:38

To Francesco: A curve on a surface is a normal crossing divisor when it has nodal singularities, thus cannot have more than two branches through each point.
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AngeloOct 4 '10 at 9:46

Oh, I see. I misread the question and I missed the requirement that $D$ should be simple normal crossing. Thank you for pointing it out, I will edit the answer.
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Francesco PolizziOct 4 '10 at 10:04